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Problem 55 Hard Difficulty

Evaluate the definite integral.

$ \displaystyle \int^1_0 \sqrt[3]{1 + 7x} \, dx $

Answer

$\frac{45}{28}$

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Video Transcript

1.7 Next to the 1/3 DX and I were looking at the integral from 01 We know we can divine are you? And then we can define our d acts. VX is do you over seven. Given this, we can now write or integral in terms of you. And now the upper limit is going to be you is one plus sometimes one, which is eight. So we now have the integral from 1 to 8. Okay, let's pull up the constant, which is 1/7. Yeah, and now we know that we can use the power rule which is increased the expert it by one and divide by the new exponents. And now we're climb a fundamental theme of calculus by plugging end our bounds to get 45 over 28.